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What is the proper equation to use when finding the tangent plane?
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My professor has introduced us to this equation when finding a tangent plane.
$$ z= f(x,y) + [frac{∂f}{∂x}(x_0,y_0,z_0)] (x-x_0) + [frac{∂f}{∂y}(x_0,y_0,z_0)] (y-y_0) $$
I am asked to find the tangent plane of the function $ f(x,y,z)= frac{xyz}{x^2+y^2+z^2}$. I get the answer $z=y$ when I solve it myself, but I saw other solutions which used the formula $$ 0 = [frac{∂f}{∂x}(x_0,y_0,z_0)] (x-x_0) + [frac{∂f}{∂y}(x_0,y_0,z_0)] (y-y_0)+[frac{∂f}{∂z}(x_0,y_0,z_0)] (z-z_0) $$
I am not sure which one to use, because this solution gives me $ y=0 $. Can anybody help clear up the difference between these two equations?
multivariable-calculus 3d
asked Jan 15 at 0:05
Justin SkaggsJustin Skaggs
31
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add a comment |
$begingroup$
My professor has introduced us to this equation when finding a tangent plane.
$$ z= f(x,y) + [frac{∂f}{∂x}(x_0,y_0,z_0)] (x-x_0) + [frac{∂f}{∂y}(x_0,y_0,z_0)] (y-y_0) $$
I am asked to find the tangent plane of the function $ f(x,y,z)= frac{xyz}{x^2+y^2+z^2}$. I get the answer $z=y$ when I solve it myself, but I saw other solutions which used the formula $$ 0 = [frac{∂f}{∂x}(x_0,y_0,z_0)] (x-x_0) + [frac{∂f}{∂y}(x_0,y_0,z_0)] (y-y_0)+[frac{∂f}{∂z}(x_0,y_0,z_0)] (z-z_0) $$
I am not sure which one to use, because this solution gives me $ y=0 $. Can anybody help clear up the difference between these two equations?
multivariable-calculus 3d
asked Jan 15 at 0:05
Justin SkaggsJustin Skaggs
31
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$begingroup$
Definitely the second one is the correct tangent plane. In the first case you have written f(x,y). What does that mean ? Because indeed f is a function on 3 variables.
$endgroup$
– Sota Antonino
Jan 15 at 0:24
add a comment |
$begingroup$
My professor has introduced us to this equation when finding a tangent plane.
$$ z= f(x,y) + [frac{∂f}{∂x}(x_0,y_0,z_0)] (x-x_0) + [frac{∂f}{∂y}(x_0,y_0,z_0)] (y-y_0) $$
I am asked to find the tangent plane of the function $ f(x,y,z)= frac{xyz}{x^2+y^2+z^2}$. I get the answer $z=y$ when I solve it myself, but I saw other solutions which used the formula $$ 0 = [frac{∂f}{∂x}(x_0,y_0,z_0)] (x-x_0) + [frac{∂f}{∂y}(x_0,y_0,z_0)] (y-y_0)+[frac{∂f}{∂z}(x_0,y_0,z_0)] (z-z_0) $$
I am not sure which one to use, because this solution gives me $ y=0 $. Can anybody help clear up the difference between these two equations?
multivariable-calculus 3d
asked Jan 15 at 0:05
Justin SkaggsJustin Skaggs
31
$endgroup$
My professor has introduced us to this equation when finding a tangent plane.
$$ z= f(x,y) + [frac{∂f}{∂x}(x_0,y_0,z_0)] (x-x_0) + [frac{∂f}{∂y}(x_0,y_0,z_0)] (y-y_0) $$
I am asked to find the tangent plane of the function $ f(x,y,z)= frac{xyz}{x^2+y^2+z^2}$. I get the answer $z=y$ when I solve it myself, but I saw other solutions which used the formula $$ 0 = [frac{∂f}{∂x}(x_0,y_0,z_0)] (x-x_0) + [frac{∂f}{∂y}(x_0,y_0,z_0)] (y-y_0)+[frac{∂f}{∂z}(x_0,y_0,z_0)] (z-z_0) $$
I am not sure which one to use, because this solution gives me $ y=0 $. Can anybody help clear up the difference between these two equations?
multivariable-calculus 3d
multivariable-calculus 3d
asked Jan 15 at 0:05
Justin SkaggsJustin Skaggs
31
asked Jan 15 at 0:05
Justin SkaggsJustin Skaggs
31
asked Jan 15 at 0:05
Justin SkaggsJustin Skaggs
31
asked Jan 15 at 0:05
Justin SkaggsJustin Skaggs
31
asked Jan 15 at 0:05
Justin SkaggsJustin Skaggs
31
31
$begingroup$
Definitely the second one is the correct tangent plane. In the first case you have written f(x,y). What does that mean ? Because indeed f is a function on 3 variables.
$endgroup$
– Sota Antonino
Jan 15 at 0:24
add a comment |
$begingroup$
Definitely the second one is the correct tangent plane. In the first case you have written f(x,y). What does that mean ? Because indeed f is a function on 3 variables.
$endgroup$
– Sota Antonino
Jan 15 at 0:24
$begingroup$
Definitely the second one is the correct tangent plane. In the first case you have written f(x,y). What does that mean ? Because indeed f is a function on 3 variables.
$endgroup$
– Sota Antonino
Jan 15 at 0:24
$begingroup$
Definitely the second one is the correct tangent plane. In the first case you have written f(x,y). What does that mean ? Because indeed f is a function on 3 variables.
$endgroup$
– Sota Antonino
Jan 15 at 0:24
add a comment |
1 Answer
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You appear to be confusing implicit and explicit functions. The second example function that you’ve given is a function from $mathbb R^3$ to $mathbb R$, so its graph doesn’t have a tangent plane, but rather a tangent hyperplane, which you can find by extending your first formula by one more dimension.
I suspect that what you are really asking about with that second function is a surface in $mathbb R^3$ that’s given implicitly by the equation $f(x,y,z)=c$ for some constant $c$. Such a surface is known as a level surface of $f$. That’s a different matter. If you can solve that equation for $z$ explicitly, i.e., describe the surface as the graph of a function of two variables, then your first formula applies. However, in many cases including this one that’s not possible, so you have to use a related, but different method.
There’s a theorem that says the gradient of a function at a point is always orthogonal to its level surface at that point. The gradient is also normal to the tangent plane (if it exists), so using the point-normal form of the equation of a plane, an equation for the tangent plane at a point $(x_0,y_0,z_0)$ on that surface is $$(x-x_0){partial foverpartial x}(x_0,y_0,z_0)+(y-y_0){partial foverpartial y}(x_0,y_0,z_0)+(z-z_0){partial foverpartial z}(x_0,y_0,z_0)=0.tag{*}$$ The two formulas are related: If you have $z=f(x,y)$, then let $g(x,y,z)=f(x,y)-z$. It should be clear that the graph of $f$ is identical to the level surface $g(x,y,z)=0$. By the chain rule we then have for the gradient $nabla g = left({partial foverpartial x},{partial foverpartial y},-1right)$ and if you plug this into (*) and rearrange you’ll end up with your first formula.
answered Jan 15 at 2:56
amdamd
30.4k21050
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1 Answer
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$begingroup$
You appear to be confusing implicit and explicit functions. The second example function that you’ve given is a function from $mathbb R^3$ to $mathbb R$, so its graph doesn’t have a tangent plane, but rather a tangent hyperplane, which you can find by extending your first formula by one more dimension.
I suspect that what you are really asking about with that second function is a surface in $mathbb R^3$ that’s given implicitly by the equation $f(x,y,z)=c$ for some constant $c$. Such a surface is known as a level surface of $f$. That’s a different matter. If you can solve that equation for $z$ explicitly, i.e., describe the surface as the graph of a function of two variables, then your first formula applies. However, in many cases including this one that’s not possible, so you have to use a related, but different method.
There’s a theorem that says the gradient of a function at a point is always orthogonal to its level surface at that point. The gradient is also normal to the tangent plane (if it exists), so using the point-normal form of the equation of a plane, an equation for the tangent plane at a point $(x_0,y_0,z_0)$ on that surface is $$(x-x_0){partial foverpartial x}(x_0,y_0,z_0)+(y-y_0){partial foverpartial y}(x_0,y_0,z_0)+(z-z_0){partial foverpartial z}(x_0,y_0,z_0)=0.tag{*}$$ The two formulas are related: If you have $z=f(x,y)$, then let $g(x,y,z)=f(x,y)-z$. It should be clear that the graph of $f$ is identical to the level surface $g(x,y,z)=0$. By the chain rule we then have for the gradient $nabla g = left({partial foverpartial x},{partial foverpartial y},-1right)$ and if you plug this into (*) and rearrange you’ll end up with your first formula.
answered Jan 15 at 2:56
amdamd
30.4k21050
$endgroup$
add a comment |
$begingroup$
You appear to be confusing implicit and explicit functions. The second example function that you’ve given is a function from $mathbb R^3$ to $mathbb R$, so its graph doesn’t have a tangent plane, but rather a tangent hyperplane, which you can find by extending your first formula by one more dimension.
I suspect that what you are really asking about with that second function is a surface in $mathbb R^3$ that’s given implicitly by the equation $f(x,y,z)=c$ for some constant $c$. Such a surface is known as a level surface of $f$. That’s a different matter. If you can solve that equation for $z$ explicitly, i.e., describe the surface as the graph of a function of two variables, then your first formula applies. However, in many cases including this one that’s not possible, so you have to use a related, but different method.
There’s a theorem that says the gradient of a function at a point is always orthogonal to its level surface at that point. The gradient is also normal to the tangent plane (if it exists), so using the point-normal form of the equation of a plane, an equation for the tangent plane at a point $(x_0,y_0,z_0)$ on that surface is $$(x-x_0){partial foverpartial x}(x_0,y_0,z_0)+(y-y_0){partial foverpartial y}(x_0,y_0,z_0)+(z-z_0){partial foverpartial z}(x_0,y_0,z_0)=0.tag{*}$$ The two formulas are related: If you have $z=f(x,y)$, then let $g(x,y,z)=f(x,y)-z$. It should be clear that the graph of $f$ is identical to the level surface $g(x,y,z)=0$. By the chain rule we then have for the gradient $nabla g = left({partial foverpartial x},{partial foverpartial y},-1right)$ and if you plug this into (*) and rearrange you’ll end up with your first formula.
answered Jan 15 at 2:56
amdamd
30.4k21050
$endgroup$
add a comment |
$begingroup$
You appear to be confusing implicit and explicit functions. The second example function that you’ve given is a function from $mathbb R^3$ to $mathbb R$, so its graph doesn’t have a tangent plane, but rather a tangent hyperplane, which you can find by extending your first formula by one more dimension.
I suspect that what you are really asking about with that second function is a surface in $mathbb R^3$ that’s given implicitly by the equation $f(x,y,z)=c$ for some constant $c$. Such a surface is known as a level surface of $f$. That’s a different matter. If you can solve that equation for $z$ explicitly, i.e., describe the surface as the graph of a function of two variables, then your first formula applies. However, in many cases including this one that’s not possible, so you have to use a related, but different method.
There’s a theorem that says the gradient of a function at a point is always orthogonal to its level surface at that point. The gradient is also normal to the tangent plane (if it exists), so using the point-normal form of the equation of a plane, an equation for the tangent plane at a point $(x_0,y_0,z_0)$ on that surface is $$(x-x_0){partial foverpartial x}(x_0,y_0,z_0)+(y-y_0){partial foverpartial y}(x_0,y_0,z_0)+(z-z_0){partial foverpartial z}(x_0,y_0,z_0)=0.tag{*}$$ The two formulas are related: If you have $z=f(x,y)$, then let $g(x,y,z)=f(x,y)-z$. It should be clear that the graph of $f$ is identical to the level surface $g(x,y,z)=0$. By the chain rule we then have for the gradient $nabla g = left({partial foverpartial x},{partial foverpartial y},-1right)$ and if you plug this into (*) and rearrange you’ll end up with your first formula.
answered Jan 15 at 2:56
amdamd
30.4k21050
$endgroup$
You appear to be confusing implicit and explicit functions. The second example function that you’ve given is a function from $mathbb R^3$ to $mathbb R$, so its graph doesn’t have a tangent plane, but rather a tangent hyperplane, which you can find by extending your first formula by one more dimension.
I suspect that what you are really asking about with that second function is a surface in $mathbb R^3$ that’s given implicitly by the equation $f(x,y,z)=c$ for some constant $c$. Such a surface is known as a level surface of $f$. That’s a different matter. If you can solve that equation for $z$ explicitly, i.e., describe the surface as the graph of a function of two variables, then your first formula applies. However, in many cases including this one that’s not possible, so you have to use a related, but different method.
There’s a theorem that says the gradient of a function at a point is always orthogonal to its level surface at that point. The gradient is also normal to the tangent plane (if it exists), so using the point-normal form of the equation of a plane, an equation for the tangent plane at a point $(x_0,y_0,z_0)$ on that surface is $$(x-x_0){partial foverpartial x}(x_0,y_0,z_0)+(y-y_0){partial foverpartial y}(x_0,y_0,z_0)+(z-z_0){partial foverpartial z}(x_0,y_0,z_0)=0.tag{*}$$ The two formulas are related: If you have $z=f(x,y)$, then let $g(x,y,z)=f(x,y)-z$. It should be clear that the graph of $f$ is identical to the level surface $g(x,y,z)=0$. By the chain rule we then have for the gradient $nabla g = left({partial foverpartial x},{partial foverpartial y},-1right)$ and if you plug this into (*) and rearrange you’ll end up with your first formula.
answered Jan 15 at 2:56
amdamd
30.4k21050
edited Jan 16 at 1:07
answered Jan 15 at 2:56
amdamd
30.4k21050
answered Jan 15 at 2:56
amdamd
30.4k21050
answered Jan 15 at 2:56
amdamd
30.4k21050
30.4k21050
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$begingroup$
Definitely the second one is the correct tangent plane. In the first case you have written f(x,y). What does that mean ? Because indeed f is a function on 3 variables.
$endgroup$
– Sota Antonino
Jan 15 at 0:24